Students can Download 2nd PUC Statistics Previous Year Question Paper March 2017, Karnataka 2nd PUC Statistics Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Statistics Previous Year Question Paper March 2017

Time: 3 Hours 15 Minutes

Max. Marks: 100

Section – A

I. Answer any Ten of the following questions : (10 × 1 = 10)

Question 1.

Define fertility

Answer:

Fertility refers to the births occurring to women of child bearing age.

Question 2.

Does Marshall – Edgeworth’s index number satisfy factor reversal test?

Answer:

Marshall – Edgeworth’s index number doesn’t satisfy factor reversal test.

Question 3.

Which price of the commodities is used in the construction of cost of living index number?

Answer:

Retail price is used in the construction of cost of living index number.

Question 4.

Give an example for seasonal variation

Answer:

The regular, periodic and short term variation (less than a year) in a time series is defined as seasonal variation.

Ex: Increase in the sale of cotton clothes in summer.

Question 5.

In which distribution mean and variance are equal?

Answer:

Poission distribution

Question 6.

Write the mean oft – distribution

Answer:

Mean = Median = Mode = 0

Question 7.

What is point estimation?

Answer:

If a single value is proposed to estimate an unknown parameter, then it is point estimation.

Question 8.

What is power of a test?

Answer:

The probability of rejecting H_{0}, when it is not true is called as power of a test. It is denoted by (1 – β).

Question 9.

When is the pooling of the frequencies done in testing of goodness of fit?

Answer:

The theoretical frequencies E_{1} ≥ 5 . If any E_{1} < 5, it should be pooled with the adjacent frequency.

Question 10.

What is defect?

Answer:

Defect: It is a quality characteristic which does not conform to specifications

Question 11.

Mention a need of replacement of an item

Answer:

- As the machine ages the cost of its maintenance increases and it requires more of care.
- If the item does not depreciate with passage of time but suddenly stops functioning.

Question 12.

What is meant by inventory?

Answer:

Inventory is a physical stock of goods kept for future use.

Section – B

II. Answer any Ten of the following (10 × 2 = 20)

Question 13.

Mention two uses of life table

Answer

- They come as a great use to the life insurance company in order to fix rates of premium for policies of different aged.
- It is used to compute the net reproduction rate and growth in population.
- It is used for future predictions.
- It is used to plan health care, retirement, education, life table serves as a important tool.
- It is used to analyze the interrelation of mortality on the basis of age, sex, composition of the population.

Question 14.

Why is Fisher’s index number called as an ‘Ideal index number’?

Answer:

Fisher’s index number is called ‘ideal index number ‘ because of the following reasons:

- It is a geometric mean which is considered as the appropriate average for averaging ratios
- It takes into account of the base year quantities as well as the current year quantities
- It is free of bias.
- It satisfies both time reversal test and factor reversal tests

Question 15.

If Σp_{1}q = 600 and Σp_{0}q = 500 then, find suitable price index number

Answer:

Kelly price index number = \(\frac{\sum p_{1} q}{\sum p_{0} q}\) × 100 = \(\frac{600}{500}\) × 100 = 120

Question 16.

What are the different phases in a business cycle?

Answer:

Prosperity, Decline, Depression, Improvement.

Question 17.

Write down the assumptions of interpolation and extrapolation?

Answer:

Assumptions of Interpolation & Extrapolation:

- There are no sudden change in the values of dependent variable from one period to another
- There is a sort of uniformity in the rise or fall of the values of the dependent variable
- There will be no Consecutive missing values in the series

Question 18.

Write down the p.m.f. of Bernoulli distribution with parameter p=0.4

Answer:

P{X) = (0.4)^{x} × (0.6)^{1-x} ,x = 0,1.

Question 19.

If Z_{1} and Z_{2} are two independent SNV’s, then name the distribution of Z_{1}^{2} + Z_{2}^{2} and find its mean

Answer:

If Z_{1}, Z_{2} are two independent SNVs, then Z_{1}^{2} + Z_{2}^{2} is chi-square distribution with 2 d.f and its mean = n = 2.

Question 20.

Define type – I and type – II errors

Answer:

Type I error is the error which occurs by rejecting null hypothesis when it is actually true

Type II error is the error which occurs by accepting null hypothesis when it is actually not true

Question 21.

If n = 10, s^{2} = 20 and σ^{2} = 25, find chi square test statistic

Answer:

χ^{2} = \(\frac{n s^{2}}{\sigma^{2}}=\frac{10 \times 20}{25}=\frac{200}{25}\) = 8

Question 22.

What do you mean by process control and product control?

Answer:

Process control: controlling the quality of the goods, during the manufacturing process itself is named as process control.

Product control: The process of inspection of manufactured lot for acceptability is named as product control or acceptance sampling.

Question 23.

If the depreciation and cumulative maintenance costs of an equipment for the third year are ₹ 10,000 and ₹ 10,400 respectively, find the annual average cost?

Answer:

\(\frac{20,400}{3}\) = 6,800

Question 24.

Given R = 3600 units/year, C_{3} = ₹ 50 and C_{1} = ₹ 4 / unit/ year, find Q^{0}

Answer:

Section – C

III. Answer any Eight of the following (8 × 5 = 40)

Question 25.

Find the standardized death rate for the following data

Answer:

Question 26.

Mention any five steps involved in the construction of index number

Answer:

The Steps involved in the construction of an index number are:

- Finding purpose and scope
- Making a survey in family budget and selection of units
- Preparing price quotations
- Calculating index numbers.

Question 27.

Find the cost of living index number for the following data

Answer:

Question 28.

Find five yearly moving averages for the following data

Answer:

Question 29.

Following is the data regarding annual net life insurance premium. Using binomial expansion method estimate the premium at the age 30 and 45

Answer:

Known values are 4

∴ (y – 1)^{4} = 0

y_{4} – 4y_{3} + 6y_{2} – 4y_{1} + y_{0} = 0 ……….. (1)

Second equation can be obtained by increasing suffixes of each term of ‘y’ by 1 and let the co-efficient same

y_{5} – 4y_{4} + 6y_{3} – 4y_{2} + y_{1} = 0 ………… (2)

Substitute y values in equation (1)

2256 – 4(1996) + 6y_{2} – 4(1581) + 1426 = 0

2256 – 7984 + 6y_{2} – 6324 + 1426 = 0

6y_{2} – 10626 = 0

y_{2} = \(\frac { 10626 }{ 6 }\)

y_{2} = 1771

Now substitute y values in equation (2)

y_{5} – 4(2256) + 6(1996) – 4(1771) + 1581 = 0

y_{5} – 9024 + 11976 – 7084 + 1581 = 0

y_{5} – 2551 = 0

y_{5} = 2551

Premium for age 30 is 1771

Premium for age 45 is 2551

Question 30.

On an average a typist makes 2 mistakes per page. Find the probability that a page typed by him has

(i) 2 mistakes

(ii) At most one mistake

Ans.

Let X=Number of mistakes λ = 2

The PMF is

P(X) = \(\frac{e^{-2} 2^{x}}{x !}\) when x = 0,1,2 …… ∞

(i) p(2 mistakes) = p(X = 2)

p(X) = \(\frac{e^{-2} \cdot 2^{2}}{2 !}\) = \(\frac{(0.1353) 4}{2}\) = 0.2706

(ii) p(at most 1 mstake) = p(X ≤ 1)

= P(0) + p(1)

p(X) = \(\frac{e^{-2} \cdot 2^{0}}{0 !}+\frac{e^{-2} \cdot 2^{1}}{1 !}\) = 0.1353 + 0.2706 = 0.4059

Question 31.

A box contains 8 red and 4 white balls from this box 5 balls are drawn at random. What is the probability that the sample contains 3 red balls

Answer:

a = 8, b = 4, n = 5,

Question 32.

Following is the data regarding mean weights of randomly selected boys and girls of PUC students. Test whether mean weight of boys is less than mean weight of girls. (Use 5% L.O.S)

Ans.

n_{1} = 64, n_{2} = 48, x̄_{1} = 50, x̄_{2} = 54, σ_{1} = 8 ,σ_{2} = 12

H_{0} = Mean weight of boys and girls is same µ_{1} = µ_{2}

H_{1} = Mean weight of boys less than girls µ_{1} < µ_{2}

It is lower tailed test

C.L = -1.65

-1.65 > -2 is rejected

Conclusion: Mean weight of boys is less than mean weight of girls

Question 33.

A specified brand of automobile tyre is known to have an average life of 40000 km. a random sample of 10 tyres of this brand, when tested resulted in the average life of 39000km and SD 1200 km. at 1% l.o.s., what is your conclusion regarding average life of tyres?

Answer:

µ = 40000km, n = 10, X̄ = 39000km, SD{s) = 1200kin

H_{0} = Average life of tyres = 40,000km µ = 40,000

H_{1} = Average life of tyres ≠ 40,000km µ ≠ 40,000

The d.f is n – 1 = 10 – 1 = 9

It is two tailed test

L0S = 1%,

CV = -2.82 and 2.82

t_{cal} value lies between the critical values

H_{0} is accepted

Conclusion: Average life of tyres is 40,000km

Question 34.

For the following data find control limits of X̄ – chart. (Given A_{2} = 0.729)

Answer:

X̿ = \(\frac{\sum \bar{X}}{n}=\frac{300}{6}\) = 50

R̄ = \(\frac{\sum R}{n}=\frac{30}{6}\) = 50

CL = X̿ = 50 = 50 – 3.645 = 46.355

UCL = X̿ + A_{2}R̄ = 50 + (0.729 × 5) = 50 + 3.645 = 53.645

Plotted points are with in limits Production is under control

Question 35.

Solve the following LPP graphically

Max Z = 100X + 170Y

s.t. 2X + 3Y ≤ 36

X + 2Y ≤ 20

And X ,Y ≥ 0

Answer:

Z = 100X + 170Y

2X + 3Y = 36

X + 2Y = 20

Equation 2X + 3Y = 36

If X = 0

2(0) + 3Y = 36 ⇒ 3Y = 36 ⇒ Y = 36/3 ⇒ Y = 12

If Y = 0

2Y + 3(0) = 36 ⇒ 2X = 36 ⇒ Y = 36/2 = 18

Points (0,12) and (18,0)

Equation X + 2Y = 20

If X = 0

0 + 2Y = 20 ⇒ 2Y = 20 ⇒ Y = 20/2 ⇒ Y = 10

If Y = 0

X + 2(0) = 20 ⇒ X = 20

Points, (0,10) and (20,0)

Corner points Z = 100X + 170Y value

A (0,10) Z = 100(0) + 170(10) = 1700

B (11,5) Z = 100(11) + 170(5) = 1950 (Max)

C (18,0) Z = 100(18) + 170(0) = 1800

0(0,0) Z = 100(0) + 170(0) = 0

LPP is maximized at (11,5)

The optimal solution is X = 11, Y = 5

Question 36.

Solve the fol1owing game by dominate rule

Answer:

Here B_{1} dominates B_{2} & B_{3}

Delete B_{2} & B_{3}

The reduce matrix is

Here A_{1} dominates A_{1}, A_{2} & A_{3}

Delete A_{1}, A_{2} & A_{3}

The reduce matrix is

Saddle point is (2,1)

The value of the game is V = 7

Section – D

IV. Answer any Two of the following questions (2 × 10 = 20)

Question 37.

Calculate GRR and NRR for the following data and comment on the result

Answer:

Question 38.

For the following data find Laspeyer’s, Paasche’s and Dorbish – Bowley’s price index numbers

Answer:

Laspeyre’s Price Index Number = \(\frac{\sum p_{1} q_{0}}{\sum p_{0} q_{0}}\) × 100 = \(\frac{228}{200}\) × 100 = 1.14 × 100 = 114

∴ 14% of the price is increased is the current year when compared to base year.

Paasche’s Price Index Number = \(\frac{\sum p_{1} q_{1}}{\sum p_{0} q_{1}}\) × 100 = \(\frac{209}{190}\) × 100 = 1.10 × 100 = 110

∴ 10% if the price is increased in the current year when compared to base year.

∴ 12% the price is increased in the current year when compared to base year.

Question 39.

For the following data fit a straight line trend and find trend values also estimate the production of 2017

Answer:

a = \(\frac{\Sigma y}{n}=\frac{630}{7}\) = 90

b = \(\frac{\sum x y}{\sum x^{2}}=\frac{84}{28}\) = 3

Trend equation y = 90 + 3x

Production for 2017 is

x = 4

y = 90 + (3) (4) = 90 + 12 = 102

Production for 2017 is 102 tons.

Question 40.

(A) Five unbiased coins are tossed 128 times. Calculate the expected frequencies for the number of heads obtained

Answer:

Let X = The number of heads obtained in toss of 5 coins

Then n = 5 p = \(\frac { 1 }{ 2 }\) q = \(\frac { 1 }{ 2 }\)

The p.m.f is p(x) = ^{5}c_{x} \(\left(\frac{1}{2}\right)^{x}\left(\frac{1}{2}\right)^{5-x}\) Where x = 0,1,2, ……… 5

The expected frequency distribution of number of number of heads is

(B) Seventy accidents that have occurred in a state in a week are tabulated as follows

Test whether accidents occurs uniformly through out the week (given K_{2} = 16.81)

Answer:

H_{0} : Accidents occur uniformly throughout the week

H_{1} : Accidents not occur uniformly throughout the week

On the basis of null hypothesis the expected frequencies are \(\frac { 70 }{ 7 }\) = 10 for all days

Applying χ^{2} – test

CV = k_{2} = 16.81

Since χ^{2}_{cal} value lies in acceptance region

Here χ^{2}_{cal} < k_{2}

H_{0} is accepted

Conclusion: accidents occur uniformly throughout the week.

Section – E

V. Answer any Two of the following (2 × 5 = 10)

Question 41.

X isa normal variate with mean 100 and SD 5. Find the probability that

(i) X ≥ 105,

(ii) 90 ≤ X ≤ 110

Answer:

Mean (µ) = 100 SD (σ) = 5

(i) P(x ≥ 105) then x = 1.5

z = \(\frac{x-\mu}{\sigma}\) = p\(\left(\frac{x-100}{5} \geq \frac{105-100}{5}\right)\)

P(Z ≥ 1) = Area from 1 tO co = 0.1587

(ii) 90 ≤ x ≤ 110

p\(\left(\frac{90-100}{5} \leq \frac{x-\mu}{\sigma} \leq \frac{110-100}{5}\right)\) = p(-2 ≤ z ≤ 2)

= (Area from -2 to ∞) – (Area from 2 to ∞)

= 0.9772 – 0.0228 = 0.9544

Question 42.

In a sample of 400 people in Kerala 220 are tea drinkers and the rest are coffee drinkers. Can we assume that both tea and coffee are equally popular in this state? Use 5% level of significance

Answer:

N = 400, x = 220, P_{0} = 0.5

H_{0} : Both coffee and tea are equally popular in state P = 0.5

H_{1} : Both coffee and tea are not equally popular in state P ≠ 0.5

It is two tailed test

α = 5%

Critical values = -1.96 & 1.96

Z_{cal} value doesn’t lie between critical values

H_{0} is rejected

Conclusion: Both coffee and tea are not equally popular in the state

Question 43.

For the following data test whether the attributes ‘smoking’ and ‘literacy’ are independent. Use 1% level of significance

Answer:

H_{0}: Smoking and literacy are independent

H_{1}: Smoking and literacy are not independent

Critical value = 6.65

χ^{2} lies in acceptance region

H_{0} is accepted

Smoking and literacy are independent.

Question 44.

For the following transportation problem find initial basic feasible solution by NWCR method. Compute transportation cost. Is the solution degenerate?

Answer:

NWCR cell is (1,1) i.e. it is 8, it’s a_{1}, b_{1} is (500, 400)

Min of a_{1}, b_{1} is 400

Delete D_{1}

Reduce matrix is

NWCR cell is (1,2) i.e. it is 4, it’s a_{1}, b_{1} is (100, 200)

Min of a_{1}, b_{1} is 100

Delete O_{1}

Reduce matrix is

NWCR cell is (2,2) i.e. it is 5, it’s a_{1}, b_{1} is (200, 100)

Min of a_{1}, b_{1} is 100

Delete D_{2}

Reduce matrix is

NWCR cell is (2,3) i.e. it is 6, it’s a_{1}, b_{1} is (100, 200)

Min of a_{1}, b_{1} is 100

Delete O_{2}

Reduce matrix is

Transportation cost is

(400 × 8) + (100 × 4) + (100 × 5) + (100 × 6) + (100 × 3)

= 3200 + 400 + 500 + 600 + 300 = 5000